Theoretically, the following 3 circuits supposed to behave equivalently, in terms of the impedance load to the voltage source at the input:

ideal transformer circuits

But to my greatest surprise, the one on the top left shows a parallel resonance within the C1-L2 tank despite the coupling coefficient is unity between L1-L2.

input currents

Is there a rational explanation for that?

  • 1
    \$\begingroup\$ It would have been nice to say why you think the three circuits are the same. It may have even avoided this question from being posed. \$\endgroup\$ Commented Sep 12, 2021 at 13:15
  • \$\begingroup\$ @a concerned citizen I began my post with that. Theoretically, the following 3 circuits supposed to behave equivalently, in terms of the impedance load to the voltage source at the input \$\endgroup\$ Commented Sep 12, 2021 at 13:28
  • \$\begingroup\$ Yes, but I meant why they do not work as you think they should. The "why" is the key. \$\endgroup\$ Commented Sep 12, 2021 at 13:33
  • \$\begingroup\$ Theoretically they do not behave equivalently. I'm unsure how you came to that conclusion. Maybe you saw it on some rogue website or maybe you misinterpreted something in a book? \$\endgroup\$
    – Andy aka
    Commented Sep 12, 2021 at 13:47
  • \$\begingroup\$ What is the response until ... 10 MHz ? \$\endgroup\$
    – Antonio51
    Commented Sep 12, 2021 at 14:29

2 Answers 2


V3 and C3 have nothing between them, they are directly coupled, which means the current will be straight differentiated. Adding a transformer will mean the capacitive load will be reflected in the primary and it will form a parallel LC tank. That's why you're seeing the notch.

Also, the bottom left circuit is reversed: the output is the VCVS and the feedback is the CCCS, while the ratio is towards the output (that's how I used it for the rest of the circuits):


I(V2) and I(V3) are identical, save for the minor shift in the phase towards DC (due to the default 1 mΩ series resistor which I left on purpose). The phase jumps are to ±180o due to the sharp discontinuity, but they are still the same since the difference is 360o.

Don't forget that the bottom left circuit is also called a "DC transformer", because while it emulates a transformer, it does so by allowing DC to pass. Therefore the output of that circuit (when drawn properly) will match the one of your V3+C3 (see I(V1)`, red trace).

Conclusion: LTspice (note the spelling) shows nothing weird, it's very rational; your assumptions are not. Maybe a different thinking is needed: if a tool that's widely used in the world by engineers all over shows a result that's not according to my thinking, I'd ask myself "what am I doing wrong?", instead of "why is the tool wrong?".

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    \$\begingroup\$ Believe me, I asked myself multiple times before posting my question here. And I did not mean to imply that anything was wrong with LTspice. I expected a pair of coupled inductors with unity coefficient to behave like an ideal transformer, i.e. to satisfy Np/Ns = Vp/Vs = Is/Ip. Although the transformer you get this way is ideal from many aspects but it is clearly not identical to the one that is commonly used in transformer equivalent circuits, where the magnetizing inductance is treated separately. This is what confused me. Anyway, thank you for making me realize what I assumed wrongly. \$\endgroup\$ Commented Sep 12, 2021 at 14:21
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    \$\begingroup\$ @AndrásGajdács You're welcome. That's why I pointed out that the DC transformer (the SPICE equivalent) is different than a coupled inductors transformer. \$\endgroup\$ Commented Sep 12, 2021 at 14:43
  • \$\begingroup\$ @AndrásGajdács Just to remember that there are some "models" ... and must be well known ... "where" and "how" must be used. Not really "wrong", just "misused". \$\endgroup\$
    – Antonio51
    Commented Sep 13, 2021 at 17:04

You appear to have forgotten about this parallel inductor (that I have added inside the red block): -

enter image description here

That inductor is L1 of course. So, at parallel resonance, the impedance will be infinite and the current will be zero. I'm uninterested in picking over your third circuit given that your analysis is flawed at the first hurdle.


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